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Kyungpook Mathematical Journal 2016; 56(2): 451-464

Published online June 1, 2016 https://doi.org/10.5666/KMJ.2016.56.2.451

Copyright © Kyungpook Mathematical Journal.

A Further Result Related to a Conjecture of R. Brück

Nan Li1, Lianzhong Yang2, Kai Liu3

School of Mathematical Sciences, University of Jinan, Jinan, Shandong, 250022, P. R. China1
School of Mathematics, Shandong University, Jinan, Shandong, 250100, P. R. China2
Department of Mathematics, Nanchang University, Nanchang, Jiangxi, 330031, P. R. China3

Received: May 9, 2013; Accepted: April 12, 2016

In this paper, we investigate the uniqueness problem of a meromorphic function sharing one small function with its differential polynomial, and give a result which is related to a conjecture of R. Brück.

Keywords: Meromorphic function, Shared value, Small function.

In this paper, meromorphic function means meromorphic in the complex plane. We use the standard notations of Nevanlinna theory, which can be found in [12]. A meromorphic function a(z) is called a small function with respect to f(z) if T(r, a) = S(r, f). We say that two meromorphic functions f and g share a small function a IM (ignoring multiplicities) when fa and ga have the same zeros. If fa and ga have the same zeros with the same multiplicities, then we say that f and g share a CM (counting multiplicities).

Let f(z) be a meromorphic function. It is known that the hyper order of f(z), denoted by σ2(f), is defined by

σ2(f)=lim suprlog log T(r,f)log r.

In 1996, R. Brück [1] posed the following conjecture.

Brück Conjecture

Let f be a non-constant entire function such that the hyper order σ2(f) of f is not a positive integer and σ2(f) < +∞. If f and f′ share a finite value a CM, then

f-af-a=c,

where c is a nonzero constant.

In 1998, Gundersen and Yang [3] verified that the Conjecture is true when f is of finite order. In 1999, Yang [10] confirmed that the Conjecture is also true when f′ is replaced by f(k)(k ≥ 2) and f is of finite order. In recent years, many results have been published concerning the above conjecture, see [2, 5, 7, 8, 14, 15, 16, 17, 18] etc., and Zhang [17] was the first author who considers the case when f is a meromorphic function. We need the following definition.

Definition 1.1

Let l be a non-negative integer or infinite. Denote by El(a, f) the set of all a-points of f where an a-point of multiplicity m is counted m times if m ≤ l and l+1 times if m > l. If El(a, f) = El(a, g), we say that f and g share (a, l).

Remark

It is easy to see that f and g share (a, l) implies that f and g share (a, p) for 0 ≤ p ≤ l. Also we note that f and g share the value a IM or CM if and only if f and g share (a, 0) or (a,∞), respectively. We also use Np(r,1f-a) to denote the counting function of the zeros of fa where a zero of multiplicity m is counted m times if m ≤ p and p times if m > p.

Lahiri [5] improved the results of Zhang [17] by using the above definition and obtained the following Theorem:

Theorem A

Let f be a non-constant meromorphic function and k be a positive integer. If f and f(k)share (1,2) and

2N¯(r,f)+N2(r,1f(k))+N2(r,1f)<(λ+o(1))T(r,f(k)),

for rI, where 0 < λ < 1 and I is a set of infinite linear measure, then f(k)-af-a=c for cC {0}.

Let p be a positive integer and aC ∪{∞}. We use Np)(r,1f-a) to denote the counting function of the zeros of fa, whose multiplicities are not greater than p, N(p+1(r,1f-a) to denote the counting function of the zeros of fa whose multiplicities are not less than p + 1, and we use N¯p)(r,1f-a) and N¯(p+1(r,1f-a) to denote their corresponding reduced counting functions (ignoring multiplicities) respectively. Define

δp(a,f)=1-lim supr+Np(r,1f-a)T(r,f).

It follows that δp(a, f) ≥ δ(a, f).

Let L(f) = f(k) + ak−1f(k−1) + · · · + a0f. Zhang and Yang [16] obtained the following result which improves the results of [5, 8, 14, 18].

Theorem B

Let f be a non-constant meromorphic function, k ≥ 1 and l ≥ 0 be integers. Let a(z) be a small function of f such that a(z) ≢ 0, ∞. Suppose that fa and L(f)−a share (0, l). Then fL(f) if one of the following assumptions holds,

  • (1) l ≥ 2 and3θ(,f)+δ2+k(0,f)+δ2(0,f)+δ(a,f)>4,

  • (2) l = 1 and7+k2θ(,f)+12δ1+k(0,f)+δ2(0,f)+δ2+k(0,f)+δ(a,f)>k2+5,

  • (3) l = 0 and(2k+6)θ(,f)+δ2(0,f)+2δ1+k(0,f)+δ2+k(0,f)+θ(0,f)+δ(a,f)>2k+10.

Definition 1.2

Let p0, p1, . . . pk be non-negative integers. We call

M[f]=fp0(f)p1(f(k))pk

a differential monomial in f with degree dM = p0 + p1 + · · · + pk and weight ΓM = p0 + 2p1 + · · · + (k + 1)pk, and

H[f]=j=1najMj[f],

where aj are small functions of f, is called a differential polynomial in f of degree d = max{dMj, 1 ≤ j ≤ n} and weight Γ = max{ΓMj, 1 ≤ j ≤ n}, furthermore if deg(Mj) = d(j = 1, 2, · · · n), then H[f] is a homogeneous differential polynomial in f of degree d.

In this paper, we improve the above Theorems and obtain the following result.

Theorem 1.3

Let f be a non-constant meromorphic function and H[f] be a non-constant homogeneous differential polynomial of degree d and weight Γ satisfying Γ (k + 2)d − 2. Let a(z) be a small meromorphic function of f such that a(z) ≢ 0, ∞. Suppose that fa and H[f] − a share (0, l). Then H[f]-af-a=C for some non-zero constant C if one of the following assumptions holds,

  • (i) l ≥ 2 and3θ(,f)+dδ2+Γ-d(0,fd)+δ2(0,f)+δ(a,f)>4,

  • (ii) l = 1 and7+Γ-d2θ(,f)+d2δ1+Γ-d(0,fd)+dδ2+Γ-d(0,fd)+δ2(0,f)+δ(a,f)>Γ+92,

  • (iii) l = 0 and[2(Γ-d)+6]θ(,f)+δ2(0,f)+dδ2+Γ-d(0,fd)+2dδ1+Γ-d(0,fd)+θ(0,f)+δ(a,f)>2Γ+8.

Especially, when l = 0, i.e. f and H share a IM, if (1.3) holds, then fH[f].

Lemma 2.1.([11])

Let f be a nonconstant meromorphic function, k be a positive integer. Then

N(r,1f(k))T(r,f(k))-T(r,f)+N(r,1f)+S(r,f),N(r,1f(k))N(r,1f)+kN¯(r,f)+S(r,f).

Suppose that F and G are two non-constant meromorphic functions such that F and G share the value 1 IM. Let z0 be a 1-point of F of order p, a 1-point of G of order q. We denote by NL(r,1F-1) the counting function of those 1-points of F where p > q, by NE1)(r,1F-1) the counting function of those 1-points of F where p = q = 1, by NE(2 the counting function of those 1-points of F where p = q ≥ 2; each point in these counting functions is counted only one time. Similarly, we can define NL(r,1G-1),NE1)(r,1G-1) and NE(2(r,1G-1).

Lemma 2.2.([13])

Let F and G are two nonconstant meromorphic functions,

Δ=(FF-2FF-1)-(GG-2GG-1).

If F and G share 1 IM and Δ ≢ 0, then

NE1)(r,1F-1)N(r,Δ)+S(r,F)+S(r,G).

Lemma 2.3

Let H[f] be a non-constant differential polynomial. Let z0be a pole of f of order p and neither a zero nor a pole of coefficients of H[f]. Then z0is a pole of H[f] with order at most pd + (Γ − d).

Proof

Let

H[f]=j=1najHj[f],         Hj[f]=fp0(f)p1(f(k))pk,dMj=p0+p1++pk,         ΓMj=p0+2p1++(k+1)pk.

Let z0 be a pole of f of order p, then z0 be a pole of Hj [f] of order pdMj + (ΓMjdMj ). Because d = max{dMj, 1 ≤ j ≤ n}, Γ = max{ΓMj, 1 ≤ j ≤ n} and z0 neither be a zero nor be a pole of aj, then z0 is a pole of H[f] with order at most pd + (Γ − d).

Lemma 2.4

Let f be a transcendental meromorphic function, H[f] is a homogeneous differential polynomial in f of degree d and weight Γ. If H[f] ≢ 0, we have

N(r,1H)T(r,H)-dT(r,f)+dN(r,1f)+S(r,f).N(r,1H)(Γ-d)N¯(r,f)+dN(r,1f)+S(r,f).
Proof

By the first fundamental theorem and the lemma of logarithmic derivatives, we have

N(r,1H)=T(r,H)-m(r,1H)+O(1)T(r,H)-(m(r,1fd)-m(r,Hfd))+O(1)=T(r,H)-(T(r,1fd)-N(r,1fd))+S(r,f)=T(r,H)-dT(r,f)+dN(r,1f)+S(r,f).

This proves (2.5). From Lemma 3, we have

T(r,H)=m(r,H)+N(r,H)m(r,Hfd)+m(r,fd)+N(r,H)dm(r,f)+dN(r,f)+(Γ-d)N¯(r,f)+S(r,f)=dT(r,f)+(Γ-d)N¯(r,f)+S(r,f).

Combining with (2.7), we obtain (2.6).

Lemma 2.5

Let f be a non-constant meromorphic function, H[f] is a homogeneous differential polynomial in f of degree d and weight Γ, and let p be a positive integer. If H[f] ≢ 0 and Γ (k + 2)d − (p + 1), we have

Np(r,1H)T(r,H)-dT(r,f)+Np+Γ-d(r,1fd)+S(r,f),Np(r,1H)(Γ-d)N¯(r,f)+Np+Γ-d(r,1fd)+S(r,f).
Proof

From (2.6), we have

Np(r,1H)+j=p+1N¯(j(r,1H)(Γ-d)N¯(r,f)+Np+Γ-d(r,1fd)+j=p+Γ-d+1N¯(j(r,1fd)+S(r,f).

Since Γ (k + 2)d − (p + 1), then we have

Np(r,1H)(Γ-d)N¯(r,f)+Np+Γ-d(r,1fd)+j=p+Γ-d+1N¯(j(r,1fd)-j=p+1N¯(j(r,1H)+S(r,f)(Γ-d)N¯(r,f)+Np+Γ-d(r,1fd)+S(r,f).

Thus (2.9) holds. By the same arguments as above, we obtain (2.8) from (2.5).

Let F=H[f]a,G=fa. From the conditions of Theorem 1.3, we know that F and G share (1, l) except the zeros and poles of a(z). From the proof of Lemma 2.4, we have

T(r,F)=O(T(r,f))+S(r,f),         T(r,G)=T(r,f)+S(r,f).

It is obvious that f is a transcendental meromorphic function. Let Δ be defined by (2.3). We distinguish two cases.

Case 1

Δ ≡ 0. Integrating (2.3), yields

1G-1=CF-1+D,

where C and D are constants and C ≠ 0. If there exists a pole z0 of f with multiplicity p which is not zero or pole of a, then z0 is a pole of F with multiplicity pd + (Γ − d), a pole of G with multiplicity p. This contradicts (3.2) as H contains at least one derivative. Therefore, we have

N¯(r,F)=N¯(r,G)=N¯(r,f)=S(r,f).

(3.2) also shows that F and G share the value 1 CM.

Next, we will prove D = 0.

Suppose D ≠ 0, then we have

1G-1=D(F-1+CD)F-1.

So, we have

N¯(r,1D(F-1+CD))=N¯(r,G-1F-1)=S(r,f).

Subcase 1.1

If CD1, then by using (3.3), (3.5) and the second fundamental theorem, we have

T(r,F)N¯(r,F)+N¯(r,1F)+N¯(r,1F-1+CD)+S(r,F)N¯(r,1F)+S(r,F)(1+o(1))T(r,F).

This gives that

T(r,F)=N¯(r,1F)+S(r,F)=N1(r,1F)+S(r,F).

So we have

T(r,H)=N(r,1H)+S(r,f)=N1(r,1H)+S(r,f).

Let p = 1, then from assumption we have

Γ(k+2)d-2=(k+2)d-(p+1).

Thus from (2.8) in Lemma 2.5, we get

T(r,H)=N1(r,1H)+S(r,f)T(r,H)-dT(r,f)+N1+Γ-d(r,1fd)+S(r,f).

So we have

dT(r,f)N1+Γ-d(r,1fd)+S(r,f).

This gives that

dT(r,f)=N1+Γ-d(r,1fd)+S(r,f).

So we have

δ2+Γ-d(0,fd)=δ1+Γ-d(0,fd)=0.

Since (3.3), we get

θ(,f)=1.

Subcase 1.1.1

l ≥ 2.

From δ2(0, f) + δ(a, f) > 1 and the definition of deficiency, we have

T(r,f)>N2(r,1f)+N(r,1f-a).

Using the second fundamental theorem of Nevanlinna and (3.3), we have

T(r,f)N¯(r,1f)+N¯(r,1f-a)+N¯(r,f)+S(r,f)=N¯(r,1f)+N¯(r,1f-a)+S(r,f).

Combining (3.7) with (3.8), we have

N2(r,1f)+N(r,1f-a)<T(r,f)N¯(r,1f)+N¯(r,1f-a)+S(r,f).

So we have

N2(r,1f)=S(r,f),N(r,1f-a)=S(r,f).

This gives that

N¯(r,1f)=S(r,f),N¯(r,1f-a)=S(r,f).

From (3.8), we get a contradiction.

Subcase 1.1.2

l = 1.

When d ≥ 2, by using (1.2) and the definition of deficiency, we get a contradiction.

When d = 1, using the similar method in subcase 1.1.1, we get a contradiction.

Subcase 1.1.3

l = 0.

By using (1.3) and the definition of deficiency, we get a contradiction.

Subcase 1.2

If CD=1, then from (3.4), we have

1G-1CFF-1.

This gives us that

(G-1-1C)F-1C.

Using that F=Ha and G=fa, we get

f-a(1+1C)-a2C·1H.

Using (3.3) (3.9), Lemma 2.3 and the first fundamental theorem, we get

(d+1)T(r,f)=T(r,1fd(f-(1+1C)a))+O(1)=T(r,-CHfda2)+O(1)=N(r,Hfd)+S(r,f)dN(r,1f)+S(r,f)(d+o(1))T(r,f),

which is a contradiction, hence D=0. This gives from (3.2) that

F-1G-1C.

So we get H[f]-af-a=C(C0).

Next, we will prove C = 1 when l = 0.

Suppose C ≠ 1, then we have

G1C(F-1+C)

and

N(r,1G)=N(r,1F-1+C).

By the second fundamental theorem and (3.3) (3.10), we have

T(r,F)N¯(r,F)+N¯(r,1F)+N¯(r,1F-1+C)+S(r,f)N¯(r,1F)+N¯(r,1G)+S(r,f)=N1(r,1F)+N¯(r,1G).

By Lemma 2.5 for p = 1, we have

dT(r,f)N1+Γ-d(r,1fd)+N¯(r,1f)+S(r,f).

From the above formula and the definition of deficiency, we have

dδ1+Γ-d(0,fd)+θ(0,f)1.

So we have

dδ2+Γ-d(0,fd)+δ2(0,f)1,dδ1+Γ-d(0,fd)1.

Combining (3.11) (3.12) (3.6) with the assumptions of Theorem 1.3, we get a contradiction.

So C = 1 and GF, i.e. fH[f].

This is just the conclusion of this theorem.

Case 2

Δ ≢ 0.

By a similar method that used in the proof of Theorem B[16], we get

T(r,F)+T(r,G)N¯(r,1F)+N¯(r,F)+N¯(r,1F-1)+N¯(r,1G)+N¯(r,G)+N¯(r,1G-1)-N0(r,1F)-N0(r,1G)+S(r,f)

and

N¯(r,1F-1)+N¯(r,1G-1)N¯(2(r,1F)+N¯(2(r,1G)+N¯(r,G)+3NL(r,1F-1)+3NL(r,1G-1)+NE1)(r,1F-1)+2NE(2(r,1G-1)+N0(r,1F)+N0(r,1G)+S(r,f).

Subcase 2.1

l ≥ 2. It is easy to see that

3NL(r,1F-1)+3NL(r,1G-1)+2NE(2(r,1G-1)+NE1)(r,1F-1)N(r,1G-1)+S(r,f).

From (3.13) (3.14) and (3.15), we have

T(r,F)+T(r,G)3N¯(r,G)+N2(r,1F)+N2(r,1G)+N(r,1G-1)+S(r,f).

Noting that

N2(r,1F)=N2(r,aH)N2(r,1H)+S(r,f).

Let p = 2, then from assumption we have

Γ(k+2)d-2>(k+2)d-(p+1).

Thus, from (2.8) in Lemma 2.5 we obtain that

T(r,H)+T(r,f)3N¯(r,f)+T(r,H)-dT(r,f)+N2+Γ-d(r,1fd)+N2(r,1f)+T(r,f)-m(r,1f-a)+S(r,f).

So we have

dT(r,f)3N¯(r,f)+N2+Γ-d(r,1fd)+N2(r,1f)-m(r,1f-a)+S(r,f).

This gives that

3θ(,f)+dδ2+Γ-d(0,fd)+δ2(0,f)+δ(a,f)4.

Which contradicts the assumption (1.1) of Theorem 1.3.

Subcase 2.2

l = 1. Noting that

2NL(r,1F-1)+3NL(r,1G-1)+2NE(2(r,1G-1)+NE1)(r,1F-1)N(r,1G-1)+S(r,f)

and

NL(r,1F-1)12N(r,FF)12T(r,FF)=12T(r,FF)+O(1)12N(r,FF)+S(r,f)12(N¯(r,1F)+N¯(r,F))+S(r,f)12(N¯(r,1H)+N¯(r,f))+S(r,f)12[(Γ-d+1)N¯(r,f)+N1+Γ-d(r,1fd)]+S(r,f).

Using the same method as Subcase 2.1, we get

dT(r,f)Γ-d+72N¯(r,f)+N2+Γ-d(r,1fd)+12N1+Γ-d(r,1fd)+N2(r,1f)-m(r,1f-a)+S(r,f).

Which contradicts with (1.2) of Theorem 1.3.

Subcase 2.3

l = 0. Noting that

NL(r,1F-1)+2NL(r,1G-1)+2NE(2(r,1G-1)+NE1)(r,1F-1)N(r,1G-1)+S(r,f).

From Lemma 2.5, we have

NL(r,1F-1)N(r,FF)N(r,FF)+S(r,f)(N¯(r,1F)+N¯(r,F))+S(r,f)(N¯(r,1H)+N¯(r,f))+S(r,f)N1+Γ-d(r,1fd)+(Γ-d+1)N¯(r,f)+S(r,f).

So we have

2NL(r,1F-1)+NL(r,1G-1)2N1+Γ-d(r,1fd)+2(Γ-d+1)N¯(r,f)+N¯(r,1f)+N¯(r,f)+S(r,f).

Combining (3.13) (3.14) (3.16) with (3.17), we have

T(r,H)+T(r,f)N2(r,1H)+N2(r,1f)+3N¯(r,f)+N(r,1f-a)+2N1+Γ-d(r,1fd)+2(Γ-d+1)N¯(r,f)+N¯(r,1f)+N¯(r,f)+S(r,f).

From (2.8), we have

N2(r,1H)T(r,H)-dT(r,f)+N2+Γ-d(r,1fd)+S(r,f).

Substituting this into (3.18), we have

dT(r,f)N2(r,1f)+2(Γ-d+3)N¯(r,f)+N¯(r,1f)+2N1+Γ-d(r,1fd)+N2+Γ-d(r,1fd)-m(r,1f-a)+S(r,f).

So we have

δ2(0,f)+θ(0,f)+2(Γ-d+3)θ(,f)+dδ2+Γ-d(0,fd)+2dδ1+Γ-d(0,fd)+δ(a,f)2Γ+8.

Which contradicts the assumption of Theorem 1.3.

Now the proof has been completed.

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